Question

In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.

Answer 1

Given: μ₀ = -1.5 cm, f₀ = 1.25 cm

We have,

`1/f_0 = 1/v_0 - 1/u_0`

`1/1.25 = 1/v_0 - 1/(-1.5)`

⇒ `v_0 = 7.5` cm

`m = v_0/u_0(1 + D/f_e)`

= `7.5/(-1.5)(1 + 25/5)`

⇒ m = -30